20t^2+29t-5=0

Solution for 20t^2+29t-5=0 equation:

20t^2+29t-5=0

a = 20; b = 29; c = -5;
Δ = b2-4ac
Δ = 292-4·20·(-5)
Δ = 1241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-\sqrt{1241}}{2*20}=\frac{-29-\sqrt{1241}}{40}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{1241}}{2*20}=\frac{-29+\sqrt{1241}}{40}
$